Chapter 5

Gases

Chemistry, Raymond Chang 10th edition, 2010

McGraw-Hill

Ahmad Aqel Ifseisi Assistant Professor of Analytical Chemistry

College of Science, Department of Chemistry

King Saud University

P.O. Box 2455 Riyadh 11451 Saudi Arabia

Building: 05, Office: 2A/149 & AA/53

Tel. 014674198, Fax: 014675992

Web site: http://fac.ksu.edu.sa/aifseisi

E-mail: ahmad3qel@yahoo.com

aifseisi@ksu.edu.sa

5.1

Substances that exist as

gases

Air is a complex mixture of several substances.

A gas is a substance that is normally in the gaseous state at ordinary

temperatures and pressures.

A vapor is the gaseous form of any substance that is a liquid or a solid at

normal temperatures and pressures.

Thus, at 25oC and 1 atm pressure, we speak of water vapor and oxygen gas.

Elements that exist as gases at 25oC and 1 atm.

The noble gases (Group 8A elements) are monatomic species; the other

elements exist as diatomic molecules. Ozone (O3) is also a gas.

Some substances found as gases at 1 atm and 25°C

All gases have the following physical characteristics:

• Gases assume the volume and shape of their containers.

• Gases are the most compressible of the states of matter.

• Gases can expand spontaneously to fill their containers.

• Gases will mix evenly and completely when confined to the same container.

• Gases have much lower densities than liquids and solids.

• Gases have relatively low molar masses.

Gases form homogeneous mixtures with each other regardless of the

identities or relative proportions of the component gases.

Characteristics of gases

5.2

Pressure of a gas

Pressure is one of the most readily measurable properties of a gas.

SI Units of Pressure

Pressure is a force applied per unit area

force = mass x acceleration

The SI unit of force is the newton (N), where

1 N = 1 kg m/s2

The SI unit of pressure is the pascal (Pa), defined as one N per m2:

1 Pa = 1 N/m2

The force experienced by any area exposed to Earth’s atmosphere is equal

to the weight of the column of air above it.

Atmospheric pressure is the pressure exerted by Earth’s atmosphere .

Atmospheric Pressure

A column of air extending from sea level to the upper atmosphere.

The actual value of atmospheric pressure

depends on location, temperature, and weather

conditions.

How is atmospheric pressure measured?

The barometer is the most familiar instrument for measuring atmospheric pressure;

consists of a long glass tube, closed at one end and filled with mercury. If the tube is

inverted in a dish of mercury so that no air enters the tube, some mercury will flow out

of the tube into the dish, creating a vacuum at the top. The weight of the mercury

remaining in the tube is supported by atmospheric pressure acting on the surface of the

mercury in the dish.

1 torr = 1 mmHg

1 atm = 760 mmHg

Standard atmospheric pressure (1 atm) is equal to the

pressure that supports a column of mercury exactly 760 mm high

at 0°C at sea level. In other words, the standard atmosphere

equals a pressure of 760 mmHg. The mmHg unit is also called

the torr.

Units of Pressure

EXAMPLE

The pressure outside a jet plane flying at high altitude falls considerably below

standard atmospheric pressure. Therefore, the air inside the cabin must be

pressurized to protect the passengers. What is the pressure in atmospheres in the

cabin if the barometer reading is 688 mmHg?

1 atm = 760 mmHg,

Practice Exercise

Convert 749 mmHg to atmospheres.

EXAMPLE

Practice Exercise

Convert 295 mmHg to kilopascals.

The atmospheric pressure in San Francisco on a certain day was 732 mmHg. What

was the pressure in kPa?

1 atm = 1.01325 x 105 Pa = 760 mmHg

A manometer is a device used to measure the pressure of gases other than the atmosphere.

The open-tube manometer is better

suited for measuring pressures equal to

or greater than atmospheric pressure.

The closed-tube manometer is normally

used to measure pressures below

atmospheric pressure.

5.3

The gas laws

Experiments with a large number of gases reveal that four variables are

needed to define the physical condition, or state of a gas:

- Temperature (T)

- Pressure (P)

- Volume (V)

- Amount of gas, usually expressed as the number of moles (n).

The equations that express the relationships among T, P, V, and n are

known as the gas laws.

-If we want to study the relationship between T & P we should fix V & n constant.

-If we want to study the relationship between P & V we should fix T & n constant.

and so on,,,

The Pressure-Volume Relationship: Boyle’s Law

The pressure of a fixed amount of gas at a constant temperature is

inversely proportional to the volume of the gas.

A mathematical expression (inverse relationship between P and V):

where k1 is the proportionality constant, V1 & V2 are the volumes at pressures P1 & P2.

The product of P & V of a gas at constant T and amount of gas is a constant.

P times V is always equal to the same constant; the value of the constant depends on T & n.

Increasing or decreasing the volume of a gas at a constant temperature.

The Temperature-Volume Relationship: Charles’s and Gay-Lussac’s Law

At constant pressure, the volume of a gas sample expands when heated and

contracts when cooled.

At any given pressure, the

plot of volume versus

temperature yields a straight

line. By extending the line to

zero volume, we find the

intercept on the temperature

axis to be -273.15oC.

In 1848 Lord Kelvin identified -273.15oC as absolute zero (0 K), theoretically the

lowest attainable temperature. Then he set up an absolute temperature scale, now

called the Kelvin temperature scale.

The volume of a fixed amount of gas maintained at constant pressure is directly

proportional to the absolute temperature of the gas.

where k2 is the proportionality constant. The equation is known as Charles’s and

Gay-Lussac’s law, or simply Charles’s law.

where V1 and V2 are the volumes of the gas at temperatures T1 and T2 (both in

kelvins), respectively.

Heating or cooling a gas at constant pressure

Heating or cooling a gas at constant volume

Another form of Charles’s law shows that at constant amount of gas and volume, the

pressure of a gas is proportional to temperature

where P1 and P2 are the pressures of the gas at temperatures T1 and T2, respectively.

The Volume-Amount Relationship: Avogadro’s Law

Dependence of volume on amount of gas at constant T and P.

Avogadro stating that at the same T and P, equal volumes of different gases

contain the same number of molecules (or atoms if the gas is monatomic).

The volume of any given gas must be proportional to the number of moles of

molecules present:

where n represents the number of moles, k4 is the proportionality constant. The

equation is the mathematical expression of Avogadro’s law; states that at constant P &

T, the volume of a gas is directly proportional to the number of moles of the gas present.

According to Avogadro’s law; when two gases react with each other, their reacting

volumes have a simple ratio to each other. If the product is a gas, its volume is

related to the volume of the reactants by a simple ratio. e.g., consider the synthesis

of ammonia from molecular hydrogen and molecular nitrogen:

3 moles 1 mole 2 moles

3 molecules 1 molecule 2 molecules

Because, at the same T & P, the volumes of gases are directly proportional to the

number of moles of the gases present, we can now write:

3 volumes 1 volume 2 volumes

The volume ratio of molecular hydrogen to molecular nitrogen is 3:1, and that of

ammonia (the product) to the sum of the volumes of molecular hydrogen and

molecular nitrogen (the reactants) is 2:4 or 1:2

5.4

The ideal gas equation

Gas laws…

We can combine all three expressions to form a single master equation for the

behavior of gases:

the ideal gas equation, describes the relationship among the four variables P, V, T, and n.

where R , the proportionality constant, is called the gas constant.

An ideal gas is a hypothetical gas whose pressure-volume-temperature

behavior can be completely accounted for by the ideal gas equation.

The molecules of an ideal gas do not attract or repel one another, and their volume

is negligible compared with the volume of the container.

Although there is no such thing in nature as an ideal gas, the ideal gas approximation

works rather well for most reasonable T and P ranges. Thus, we can safely use the

ideal gas equation to solve many gas problems.

Before we can apply the ideal gas

equation to a real system, we must

evaluate the gas constant R.

PV nT

R =

In working problems with the ideal gas equation, the units of P, T, n and V must

agree with the unit in the gas constant

At 0oC (273.15 K) and 1 atm pressure, many real gases behave like an ideal

gas. These conditions are called standard temperature and pressure (STP).

Suppose you have 1.00 mol of an ideal gas at STP. According to the ideal gas

equation, the volume of the gas is:

1 mole of an ideal gas occupies 22.414 L,

which is somewhat greater than the volume

of a basketball.

The volume occupied by one mole of ideal gas at STP (22.41 L) is known as the

molar volume of an ideal gas at STP.

EXAMPLE

Sulfur hexafluoride (SF6) is a colorless, odorless, very unreactive gas.

Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel

vessel of volume 5.43 L at 69.5oC.

Practice Exercise

Calculate the volume (in liters) occupied by 2.12 moles of nitric oxide (NO) at

6.54 atm and 76oC.

Calculate the volume (in liters) occupied by 7.40 g of NH3 at STP.

EXAMPLE

By using the fact that the molar volume of a gas occupies 22.41 L at STP, we can

calculate the volume of a gas at STP without using the ideal gas equation.

nRT P

V =

Practice Exercise

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

n = 7.4 / 17.03 = 0.4345 mol

= (0.4345 x 0.0821 x 273) / 1 = 9.74 L

If we need to deal with changes in pressure, volume, temperature, or the amount of

gas, we must employ a modified form of the ideal gas equation that takes into

account the initial and final conditions.

If n1 = n2, as is usually the case because the amount of gas normally does not

change, the equation then becomes

The subscripts 1 and 2 denote the initial and final

states of the gas, respectively.

EXAMPLE

An inflated helium balloon with a volume of 0.55 L at sea level (1.0 atm) is allowed

to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that

the temperature remains constant, what is the final volume of the balloon?

Because n1 = n2 and T1 = T2,

Therefore,

Practice Exercise

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg.

Calculate the pressure of the gas (in mmHg) if the volume is reduced at constant

temperature to 154 mL.

As pressure decreased, volume increased

EXAMPLE

Argon is an inert gas used in lightbulbs to retard the vaporization of the tungsten

filament. A certain lightbulb containing argon at 1.20 atm and 18oC is heated to 85oC

at constant volume. Calculate its final pressure (in atm).

Because n1 = n2 and V1 = V2, the Equation becomes

Practice Exercise

A sample of oxygen gas initially at 0.97 atm is cooled from 21oC to -68oC at constant

volume. What is its final pressure (in atm)?

EXAMPLE

A small bubble rises from the bottom of a lake, where the T and P are 8oC and 6.4 atm,

to the water’s surface, where the temperature is 25oC and the pressure is 1.0 atm.

Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL.

the amount of air in the bubble remains constant, that is,

n1 = n2 so that

Practice Exercise

A gas initially at 4.0 L, 1.2 atm, and 66oC undergoes a change so that its final

volume and temperature are 1.7 L and 42oC. What is its final pressure? Assume the

number of moles remains unchanged.

Density Calculations

If we rearrange the ideal gas equation, we can calculate the density of a gas:

PV = nRT P M

RT d =

Unlike molecules in condensed matter (that is, in liquids and solids), gaseous

molecules are separated by distances that are large compared with their size.

Consequently, the density of gases is very low under atmospheric conditions. For

this reason, gas densities are usually expressed in (g/L) rather than (g/mL).

n V

P RT

= 𝒏 =

𝒎

M

𝒅 =𝒎

V

EXAMPLE

Calculate the density of carbon dioxide (CO2) in grams per liter (g/L) at

0.990 atm and 55oC.

Practice Exercise

What is the density (in g/L) of uranium hexafluoride (UF6) at 779 mmHg

and 62oC?

T = 273 + 55 = 328 K

The Molar Mass of a Gaseous Substance

P M

RT d =

dRT

P M =

An apparatus for measuring

the density of a gas

A bulb of known volume is filled with the gas under

study at a certain T and P. First the bulb is weighed,

and then it is emptied (evacuated) and weighed again.

The difference in masses gives the mass of the gas.

Knowing the volume of the bulb, we can calculate the

density of the gas. Once we know the density of a gas,

we can calculate the molar mass of the substance.

EXAMPLE

A chemist has synthesized a greenish-yellow gaseous compound of chlorine and

oxygen and finds that its density is 7.71 g/L at 36oC and 2.88 atm. Calculate the

molar mass of the compound and determine its molecular formula.

Alternatively, the compound must contain one Cl atom

and two O atoms and have the formula

ClO2, which has a molar mass of 67.45 g.

EXAMPLE

Chemical analysis of a gaseous compound showed that it contained 33.0 percent silicon

(Si) and 67.0 percent fluorine (F) by mass. At 35oC, 0.210 L of the compound exerted a

pressure of 1.70 atm. If the mass of 0.210 L of the compound was 2.38 g, calculate the

molecular formula of the compound.

By assuming that we have 100 g of the compound, so the percentages are converted to

grams. The number of moles of Si and F are:

The empirical formula is Si1.17F3.53, dividing by

the smaller subscript (1.17), we obtain SiF3.

The molar mass of the empirical formula SiF3 is 85.09 g.

The ratio (molar mass/empirical molar mass) is always an integer (169/85.09 ~ 2).

Therefore, the molecular formula of the compound must be (SiF3) 2 or Si2F6.

Practice Exercise

A gaseous compound is 78.14 percent boron and 21.86

percent hydrogen. At 27°C, 74.3 mL of the gas exerted a

pressure of 1.12 atm. If the mass of the gas was 0.0934 g,

what is its molecular formula?

Practice Exercise

The density of a gaseous organic compound is 3.38 g/L at

40oC and 1.97 atm. What is its molar mass?

5.5

Gas stoichiometry

Calculate the volume of O2 (in liters) required for the complete combustion of 7.64 L

of acetylene (C2H2) measured at the same temperature and pressure.

EXAMPLE

According to Avogadro’s law, at the same T and P, the number of moles of gases

are directly related to their volumes.

Therefore, we can also write

From the equation,

EXAMPLE

Sodium azide (NaN3) is used in some automobile air bags. The impact of a collision

triggers the decomposition of NaN3 as follows:

The nitrogen gas produced quickly inflates the bag between the driver and the

windshield and dashboard. Calculate the volume of N2 generated at 80°C and 823

mmHg by the decomposition of 60.0 g of NaN3.

EXAMPLE

Aqueous lithium hydroxide solution is used to purify air in spacecrafts and submarines

because it absorbs carbon dioxide, which is an end product of metabolism, according to

the equation

The pressure of carbon dioxide inside the cabin of a submarine having a volume of 2.4x105 L

is 7.9x10-3 atm at 312 K. A solution of lithium hydroxide (LiOH) of negligible volume is

introduced into the cabin. Eventually the pressure of CO2 falls to 1.2x10-4 atm. How many

grams of lithium carbonate Li2CO3 are formed by this process?

The drop in CO2 pressure is (7.9x10-3 atm) - (1.2x10-4 atm) = 7.8 x 10-3 atm.

Therefore, the number of moles of CO2 reacted is given by

so the amount of Li2CO3 formed is also 73 moles.

Practice Exercise

The equation for the metabolic breakdown of glucose (C6H12O6) is the same as the

equation for the combustion of glucose in air:

Calculate the volume of CO2 produced at 37°C and 1.00 atm when 5.60 g of

glucose is used up in the reaction.

Practice Exercise

A 2.14-L sample of hydrogen chloride (HCl) gas at 2.61 atm and 28°C is completely

dissolved in 668 mL of water to form hydrochloric acid solution. Calculate the

molarity of the acid solution. Assume no change in volume.

Practice Exercise

Assuming no change in temperature and pressure, calculate the volume of O2 (in

liters) required for the complete combustion of 14.9 L of butane (C4H10)

5.6

Dalton's law of partial

pressures

How do we deal with gases composed of a mixture of two or more substances ???

- Dalton’s Law of Partial Pressures

The total pressure of a mixture of gases equals the sum of the pressures

that each would exert if it were present alone.

Ptotal = P1 + P2 + P3 + …

The pressure exerted by a particular component of a mixture of gases is called the

partial pressure of that gas.

Consider a case in which two gases, A and B, are in a container of volume V. The

pressure exerted by gases A and B, according to the ideal gas equation, is:

In a mixture of gases A and B, the total pressure PT is the result of the collisions of both

types of molecules, A and B, with the walls of the container. Thus, according to Dalton’s law,

where n, the total number of moles of gases

present, is given by n = nA + nB, and PA and PB are

the partial pressures of gases A and B, respectively.

For a mixture of gases, then, PT depends only on the total number of moles of gas

present, not on the nature of the gas molecules.

In general, the total pressure of a mixture of gases is given by

where P1, P2, P3 , . . . are the partial pressures of components 1, 2, 3, . . . .

To see how each partial pressure is related to the total pressure, consider again the

case of a mixture of two gases A and B. Dividing PA by PT, we obtain

where XA is called the mole fraction of A.

The mole fraction is a dimensionless quantity that expresses the ratio of the number

of moles of one component to the number of moles of all components present.

where ni and nT are the number of moles of component i and

the total number of moles present, respectively.

The mole fraction is always smaller than 1. We can now express the partial

pressure as:

The sum of the mole fractions for a mixture of gases must be unity (1). If only two

components are present, then

If a system contains more than two gases, then the partial pressure of the i th

component is related to the total pressure by:

EXAMPLE

A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole of argon (Ar), and

2.15 moles of xenon (Xe). Calculate the partial pressures of the gases if the total

pressure is 2.00 atm at a certain temperature.

(1.21 + 0.20 + 0.586) atm = 2.00 atm.

When one collects a gas over water, there is water vapor mixed in with the gas.

-To find only the pressure of the desired gas, one must subtract the vapor pressure

of water from the total pressure.

Collecting Gases Over Water

An experiment that is often encountered in general chemistry laboratories involves

determining the number of moles of gas collected from a chemical reaction.

Sometimes this gas is collected over water.

The oxygen gas can be

collected over water.

Oxygen gas generated by the decomposition of potassium chlorate is collected over

water. The volume of oxygen collected at 24°C and atmospheric pressure of 762

mmHg is 128 mL. Calculate the mass (in grams) of oxygen gas obtained. The

pressure of the water vapor at 24°C is 22.4 mmHg.

EXAMPLE

From the ideal gas equation

Practice Exercise

A sample of natural gas contains 8.24 moles of methane (CH4), 0.421 mole

of ethane (C2H6), and 0.116 mole of propane (C3H8). If the total pressure of

the gases is 1.37 atm, what are the partial pressures of the gases?

Practice Exercise

Hydrogen gas generated when calcium metal reacts with water is collected

over water. The volume of gas collected at 30°C and pressure of 988

mmHg is 641 mL. What is the mass (in grams) of the hydrogen gas

obtained? The pressure of water vapor at 30°C is 31.82 mmHg.

5.7

The kinetic molecular

theory of gases

The Kinetic Molecular Theory of Gases

The kinetic molecular theory of gases assumptions:

1. A gas is composed of molecules that are separated from each other by distances far

greater than their own dimensions. The molecules can be considered to be “points”; that

is, they possess mass but have negligible volume.

2. Gas molecules are in constant motion in random directions, and they frequently collide

with one another. Collisions among molecules are perfectly elastic. In other words,

energy can be transferred from one molecule to another as a result of a collision.

Nevertheless, the total energy of all the molecules in a system remains the same.

3. Gas molecules exert neither attractive nor repulsive forces on one another.

4. The average kinetic energy of the molecules is proportional to the temperature of the

gas in kelvins. Any two gases at the same temperature will have the same average

kinetic energy. The average kinetic energy of a molecule is given by:

where: m is the mass of the molecule and u is its speed.

The horizontal bar denotes an average value. KE = ½ m u2

According to the kinetic molecular theory, gas pressure is the result of collisions between

molecules and the walls of their container. It depends on the frequency of collision per unit area

and on how “hard” the molecules strike the wall.

Application to the Gas Laws

Compressibility of Gases …

Because molecules in the gas phase are separated by large distances,

gases can be compressed easily to occupy less volume.

Boyle’s Law …

P a collision rate with wall

Collision rate a number density

Number density a 1/V P a 1/V

Charles’s Law …

P a collision rate with wall

Collision rate a average kinetic energy of gas molecules

Average kinetic energy a T P a T

Avogadro’s Law …

P a collision rate with wall

Collision rate a number density

Number density a n P a n

Dalton’s Law of Partial Pressures …

Molecules do not attract or repel one another

P exerted by one type of molecule is unaffected by the

presence of another gas Ptotal = SPi

Distribution of Molecular Speeds

The distribution of speeds for nitrogen gas at three

different temperatures. At the higher temperatures, more

molecules are moving at faster speeds.

The distribution of speeds for three gases at 300 K. At a

given temperature, the lighter molecules are moving

faster, on the average.

The distribution curve tells us the number of molecules moving at a certain speed.

The peak of each curve represents the most probable speed, that is, the speed of

the largest number of molecules.

Root-Mean-Square Speed

Root-mean-square (rms) speed (urms) is an average molecular speed.

urms = 3RT M

urms: root-mean-square speed (m/s)

R = 8.314 J/K.mol M: molar mass (kg/mol)

One of the results of the kinetic theory of gases is that the total kinetic

energy of a mole of any gas = 3

2RT.

The average kinetic energy of one molecule = 𝟏

𝟐 mu2

The heavier the gas, the more slowly its molecules move.

EXAMPLE

Calculate the root-mean-square speeds of helium atoms and nitrogen molecules in

m/s at 25°C.

Solution:

For He: The molar mass of He is 4.003 g/mol, or 4.003x10-3 kg/mol.

For N2: the molar mass is 28.02 g/mol, or 2.802x10-2 kg/mol

Using the conversion factor:

1 J = 1 kg m2/s2

Practice Exercise

Calculate the root-mean-square speed of molecular chlorine in m/s at 20°C.

Gas Diffusion and Effusion

Gas Diffusion The gradual mixing of molecules of one gas with molecules of another by virtue of

their kinetic properties.

Thomas Graham found that under the same

conditions of temperature and pressure, rates of

diffusion for gases are inversely proportional to

the square roots of their molar masses.

Graham’s law of diffusion

where r 1 and r 2 are the diffusion rates of gases 1 and 2, and M1 and M2 are

their molar masses, respectively.

r1

r2

M2 M1 =

The path traveled by a single gas molecule.

Each change in direction represents a

collision with another molecule.

A demonstration of gas diffusion. NH3 gas (from a

bottle containing aqueous ammonia) combines with

HCl gas (from a bottle containing hydrochloric acid)

to form solid NH4Cl. Because NH3 is lighter and

therefore diffuses faster, solid NH4Cl first appears

nearer the HCl bottle (on the right).

HCl: 36 g/mol

NH3: 17 g/mol

Gas Effusion

The process by which a gas under pressure escapes from one compartment of a

container to another by passing through a small opening.

NH3 HCl

Although effusion differs from diffusion in nature,

the rate of effusion of a gas has the same form as

Graham’s law of diffusion.

= r1

r2

t2

t1

M2 M1 =

where t1 and t2 are

the times for effusion

for gases 1 and 2,

respectively.

EXAMPLE

A flammable gas made up only of carbon and hydrogen is found to effuse through

a porous barrier in 1.50 min. Under the same conditions of temperature and

pressure, it takes an equal volume of bromine vapor 4.73 min to effuse through the

same barrier. Calculate the molar mass of the unknown gas, and suggest what this

gas might be.

Solution:

From the molar mass of Br2

Because the molar mass of carbon is 12.01 g and that of hydrogen is 1.008 g, the

gas is methane (CH4).

Practice Exercise

It takes 192 s for an unknown gas to effuse through a porous wall and 84 s for the

same volume of N2 gas to effuse at the same temperature and pressure. What is

the molar mass of the unknown gas?

EXAMPLE

Nickel forms a gaseous compound of the formula Ni(CO)x What is the value of x

given that under the same conditions methane (CH4) effuses 3.3 times faster than

the compound?

r1 = 3.3 x r2 M2 = r1

r2 ( )

2 x M1

= (3.3)2 x 16 = 174.2

58.7 + (x • 28) = 174.2

x = 4.1 ~ 4

Solution

The molar mass of CH4 is 16 g and that of Ni is 58.7 g

r1

r2

M2 M1 =

5.8

Deviation from ideal

behavior

Deviations from Ideal Behavior

Although we can assume that real gases behave like an ideal gas, we

cannot expect them to do so under all conditions.

Significant deviations occur as

pressure increases or temperature

decrease.

According to the ideal gas equation,

for 1 mole of gas,

PV / RT = 1

regardless of the actual gas

pressure and temperature.

Van der Waals equation nonideal gas

Effect of intermolecular forces on

the pressure exerted by a gas.

Having taken into account the corrections for

pressure and volume, we can rewrite the ideal

gas equation as follows:

van der Waals constants

of some common gases

The value of a indicates how strongly

molecules of a given type of gas attract one

another. Helium atoms have the weakest

attraction for one another, because helium

has the smallest a value.

There is also a rough correlation between

molecular size and b. Generally, the larger

the molecule (or atom), the greater b is, but

the relationship between b and molecular (or

atomic) size is not a simple one.

EXAMPLE

Given that 3.50 moles of NH3 occupy 5.20 L at 47°C, calculate the pressure of the

gas (in atm) using (a) the ideal gas equation and (b) the van der Waals equation.

Solution:

V = 5.20 L, T = (47 + 273) K = 320 K, n = 3.50 mol, R = 0.0821 L.atm/K.mol

(a) Using the ideal gas equation,

(b) Using the van der Waals equation,

For NH3;

a = 4.17 atm.L2/mol2, b = 0.0371 L/mol

Practice Exercise

Using the data shown in the van der Waals constants Table, calculate the pressure

exerted by 4.37 moles of molecular chlorine confined in a volume of 2.45 L at 38°C.

Compare the pressure with that calculated using the ideal gas equation.